Cho tích phân từ 1 đến 2 của can(1+x^2)/ x^4 dx=1/c (acana- bcanb/ b+c)

* Hướng dẫn giải

Đáp án A.

$$\begin{gathered} \mathrm{Đặt} \mathrm{x}=\mathrm{tant}\iff \left\{\begin{array}{l}1+{\mathrm{x}}^2=1+{\tan }^2\mathrm{t}=\frac{1}{{\cos }^2\mathrm{t}}\\ \mathrm{dx}=\frac{1}{{\cos }^2\mathrm{t}}\mathrm{dt}\end{array}\right. \\ \mathrm{Đổi} \mathrm{cận}: \mathrm{x}=1\Rightarrow \mathrm{t}=\frac{\mathrm{\pi }}{4}; \mathrm{x}=2\Rightarrow \mathrm{t}=\arctan 2 \\ \Rightarrow \mathrm{I}={\int }_{\frac{\mathrm{\pi }}{4}}^{\arctan 2}\frac{1}{\mathrm{cost}.{\left(\mathrm{tant}\right)}^2}.\frac{1}{{\cos }^2\mathrm{t}}\mathrm{dt} \\ \mathrm{I}={\int }_{\frac{\mathrm{\pi }}{4}}^{\arctan 2}{\cot }^4\mathrm{t}.\frac{1}{{\cos }^3\mathrm{t}}\mathrm{dt}={\int }_{\frac{\mathrm{\pi }}{4}}^{\arctan 2}\frac{\mathrm{cost}}{{\sin }^4\mathrm{t}}\mathrm{dt} \\ \mathrm{Đặt} \mathrm{u}=\mathrm{sint}\Rightarrow \mathrm{du}=\mathrm{costdt} \\ \mathrm{Đổi} \mathrm{cận}: \mathrm{t}=\frac{\mathrm{\pi }}{4}\Rightarrow \mathrm{u}=\frac{\sqrt{2}}{2}; \mathrm{t}=\arctan 2\Rightarrow \mathrm{u}=\frac{2\sqrt{5}}{5} \\ \Rightarrow \mathrm{I}={\int }_{\frac{\sqrt{2}}{2}}^{\frac{2\sqrt{5}}{5}}\frac{\mathrm{du}}{{\mathrm{u}}^4}={\overline{)\frac{-1}{3{\mathrm{u}}^3}} }_{\frac{\sqrt{2}}{2}}^{\frac{2\sqrt{5}}{5}}=\frac{-5\sqrt{5}}{24}+\frac{2\sqrt{2}}{3} \\ =\frac{1}{3}\left(2\sqrt{2}-\frac{5\sqrt{5}}{8}\right) \\ \Rightarrow \mathrm{c}=3; \mathrm{a}=2; \mathrm{b}=5 \\ \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=2+5+3=10 \end{gathered}$$